§ 155. (1) A Jet impinges on a plane surface at rest, in a direction normal to the plane (fig. 154).—Let a jet whose section is ω impinge with a velocity v on a plane surface at rest, in a direction normal to the plane. The particles approach the plane, are gradually deviated, and finally flow away parallel to the plane, having then no velocity in the original direction of the jet. The quantity of water impinging per second is ωv. The pressure on the plane, which is equal to the change of momentum per second, is P = (G/g) ωv2.
(2) If the plane is moving in the direction of the jet with the velocity ±u, the quantity impinging per second is ω(v ± u). The momentum of this quantity before impact is (G/g)ω(v ± u)v. After impact, the water still possesses the velocity ±u in the direction of the jet; and the momentum, in that direction, of so much water as impinges in one second, after impact, is ±(G/g) ω (v ± u)u. The pressure on the plane, which is the change of momentum per second, is the difference of these quantities or P = (G/g) ω (v ± u)2. This differs from the expression obtained in the previous case, in that the relative velocity of the water and plane v ± u is substituted for v. The expression may be written P = 2 × G × ω (v ± u)2/2g, where the last two terms are the volume of a prism of water whose section is the area of the jet and whose length is the head due to the relative velocity. The pressure on the plane is twice the weight of that prism of water. The work done when the plane is moving in the same direction as the jet is Pu = (G/g) ω (v − u)2u foot-pounds per second. There issue from the jet ωv cub. ft. per second, and the energy of this quantity before impact is (G/2g) ωv3. The efficiency of the jet is therefore η = 2(v − u)2u/v3. The value of u which makes this a maximum is found by differentiating and equating the differential coefficient to zero:—
dη / du = 2 (v2 − 4vu + 3u2) / v3 = 0;
∴ u = v or 1⁄3 v.
The former gives a minimum, the latter a maximum efficiency.
Putting u = 1⁄3v in the expression above,
η max. = 8⁄27.
(3) If, instead of one plane moving before the jet, a series of planes are introduced at short intervals at the same point, the quantity of water impinging on the series will be ωv instead of ω(v − u), and the whole pressure = (G/g) ωv (v − u). The work done is (G/g)ωvu (v − u). The efficiency η = (G/g) ωvu (v − u) ÷ (G/2g) ωv3 = 2u(v-u)/v2. This becomes a maximum for dη/du = 2(v − 2u) = 0, or u = 1⁄2v, and the η = 1⁄2. This result is often used as an approximate expression for the velocity of greatest efficiency when a jet of water strikes the floats of a water wheel. The work wasted in this case is half the whole energy of the jet when the floats run at the best speed.
§ 156. (4) Case of a Jet impinging on a Concave Cup Vane, velocity of water v, velocity of vane in the same direction u (fig. 155), weight impinging per second = Gw (v − u).
| Fig. 155. |